A rectangle inside a circular sector of an acute angle

Question

In the math puzzle book by Gardner, the maximum area ${\cal A}$ of the rectangle inscribed in the sector of unit-radius circle (see the Fig.) with angle $0 < \theta \le \pi/2$ has been asked to show as $${\cal A }=\frac{1-\cos \theta}{2\sin \theta}~~~~~~(1).$$ By considering point B critically on the arc along the angle bisector OB, we can prove the maximal result (1) as:

If O is the origin then $OD=AD \cot \theta$, $AD=BC=\sin(\theta/2)$, $OC=\cos(\theta/2)$. Max area ${\cal A}$ of the rectangle $${\cal A}= BC.DC=BC.[OC-OD]= \sin(\theta/2)[\cos(\theta/2)-\sin(\theta/2).\cot \theta]$$ $$=[\sin(\theta/2)\cos(\theta/2)-\sin^2(\theta/2)\cot(\theta)]=\frac{1}{2}[ \sin \theta-(1-\cos\theta) \cot \theta]$$ $$=\frac{1}{2}\left(\frac{\sin^2\theta-\cos\theta +\cos^2 \theta}{\sin\theta}\right)=\frac{1-\cos \theta}{2 \sin \theta},~~ 0<\theta \le \pi/2.$$ The question is as to what are other proofs of (1) given a fixed acute angle $\theta$.

Answer 1

Let $\angle BOC=u$

Area = $$\sin u ( \cos u- \sin u \cot \theta)$$

Differentiate with respect to $u$ and simplify

$$ \dfrac {\sin u }{ ( \cos u- \sin u \cot \theta) } =- \dfrac {\cos u}{\sin u+ \cos u \cot \theta} $$ $$ \cot \theta = \cot 2 u $$

$$ u=\dfrac{\theta}{2}$$ In other words $OB$ is the bisector of sector angle for maximum rectangle area. Since $\theta$ is assumed constant, it is bisector of any chosen sector angle.

Next for second query plug this into Area and simplify $$ \dfrac12 \tan (\theta/2)$$

which equals (1), answer is yes globally, meaning any acute angle chosen for sector.

Answer 2

Let $\angle BOC=\phi$, $|AB|=|CD|=a(\phi),\ |BC|=|AD|=b(\phi)$.

Then

\begin{align} b(\phi)&=\sin\phi ,\\ a(\phi)&=\cos\phi-\sin\phi\cot\theta ,\\ S(\phi)&=a(\phi)b(\phi) =(\cos\phi-\sin\phi\cot\theta)\,\sin\phi ,\\ S'(\phi)&= \cos^2\phi-\sin^2\phi-2\sin\phi\cos\phi\,\cot\theta =\cos2\phi-\sin2\phi\,\cot\theta . \end{align}

$S'(\phi)=0$ results in

\begin{align} \cos2\phi&=\sin2\phi\,\cot\theta ,\\ \cot2\phi&=\cot\theta ,\\ \phi&=\tfrac12\,\theta ,\\ \text{for }\theta\in(0,\tfrac\pi2)\quad \max_{\phi\in(0,\theta)} S(\phi) &= S(\tfrac12\,\theta) =\frac{1-\cos\theta}{2\,\sin\theta} . \end{align}

Answer 3

Let the unit-radius sector have measure $2\alpha$, and let the vertex of the inscribed rectangle determine an angle $\alpha-\phi$ (where $\phi$ could be negative), as shown:

Then the area of the rectangle is given by

$$\begin{align} &\;\sin(\alpha-\phi)\left(\cos(\alpha-\phi)-\sin(\alpha-\phi)\cot2\alpha \right) \\[4pt] =&\;\frac{\sin(\alpha-\phi)}{\sin2\alpha}\left(\sin2\alpha\cos(\alpha-\phi)-\cos2\alpha\sin(\alpha-\phi)\right) \\[4pt] =&\;\frac{\sin(\alpha-\phi)\sin(\alpha+\phi)}{\sin2\alpha} \\[4pt] =&\;\frac{\cos 2\phi - \cos 2\alpha}{2\sin2\alpha} \end{align}$$ which is clearly maximized when $\phi=0$. $\square$

Answer 4

Let the co-ordinate of A be (x,y) and then that of B is $(\sqrt{1-y^2},y)$ So the area of ABCD is given by $$a(y)=y(\sqrt{1-y^2}-y \cot \theta)~~~~(1)$$ Setting $a'(y)=0$ for max, we get $$a'(y)=\sqrt{1-y^2}+y\frac{-2y}{2\sqrt{1-y^2}}-2\cot \theta=0 \implies 1-2y^2=2y\sqrt{1-y^2} \cot \theta $$ $$\implies 4 \csc^2 \theta~ y^4-4 \csc^2 \theta ~y^2+1=0$$ Discarding $+$ sign in the root as it gives $y(0)=\ne 0$, we get $y^2=\frac{ 1- \cos \theta}{2}$, putting this in (1), we get $${\cal A}= \frac{1-\cos \theta}{2 \sin \theta}.$$

Answer 5

EDIT1:

As to your supplementary question about global, it was felt that a separate question may be better.

However it is clear that the area fraction Rectangle/ Sector does not change when we consider the rectangles symmetrical about previous sector's base. In other words, bisecting the sector into two parts and then again bisecting each half into two, making four equal pies furnishes the combination solution below needing no new proof.

$$ \dfrac{A_{rect 1}}{A_{sector 1}}+\dfrac{A_{rect 2}}{A_{sector 2}}=\dfrac{A_{rect}}{A_{sector}}$$

As the areas are symmetrically disposed, maximum fraction of area is ensured.

Answer 6

Let $\alpha = \angle BOC$. Then, the area is

\begin{align} Area & = AD \cdot AB = \sin\alpha\cdot (\cos\alpha - \sin\alpha \cot \theta) \\ & = \frac{\cos(2\alpha -\theta)}{2\sin\theta} - \frac12\cot\theta\\ & \le \frac{1}{2\sin\theta} - \frac12\cot\theta = \frac{1-\cos\theta}{2\sin\theta} \end{align}

where the maximum area occurs at $\cos(2\alpha - \theta)=1$, or $\alpha = \frac12\theta$.